Math 22 Lagrange Multipliers

Method of Lagrange Multipliers

 If $f(x,y)$ has a maximum or minimum subject to the constraint $g(x,y)=0$, then it will occur at one of the critical numbers of the function  defined by
 $F(x,y,\lambda )=f(x,y)-\lambda g(x,y)$.
 
 In this section, we need to set up the system of equations:
 
 $F_{x}(x,y,\lambda )=0$
 $F_{y}(x,y,\lambda )=0$
 $F_{\lambda }(x,y,\lambda )=0$

Example: Set up the Lagrange Multipliers:

1) Maximum: $f(x,y)=xy$ and Constraint $x+y-14=0$

Solution:
So, $F(x,y,\lambda )=f(x,y)-\lambda g(x,y)=xy-\lambda (x+y-14)=xy-\lambda x-\lambda y+14\lambda$
$F_{x}(x,y,\lambda )=y-\lambda$
$F_{y}(x,y,\lambda )=x-\lambda$
$F_{\lambda }(x,y,\lambda )=-x-y+14$

2) Maximum: $f(x,y)=xy$ and Constraint $x+3y-6=0$

Solution:
So, $F(x,y,\lambda )=f(x,y)-\lambda g(x,y)=xy-\lambda (x+3y-6)=xy-\lambda x-3\lambda y-6\lambda$
$F_{x}(x,y,\lambda )=y-\lambda$
$F_{y}(x,y,\lambda )=x-3\lambda$
$F_{\lambda }(x,y,\lambda )=-x-3y+6$

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